19. Properties of Triangles’ Solutions Part 2

51. The diagram is given below:

    

    From figure, $\angle ADC = 90^\circ + B$

    By applying $m:n$ rule in triangle $ABC,$ we get

    $(1 + 1)\cot(90^\circ + B) = 1.\cot90^\circ - 1.\cot(A-90^\circ)$

    $-2\tan B = 0 - \cot[-(90^\circ - A)]$

    $-2\tan B = \tan A \Rightarrow \tan A + 2\tan B = 0$

52. Given, $\cot A + \cot B + \cot C = \sqrt{3}$

    Squaring, $\cot^2A + \cot^2B + \cot^2C + 2\cot A\cot B + 2\cot B\cot C + 2\cot C\cot A = 3$

    Since $A + B + C = \pi \Rightarrow A + B = \pi - C$

    $\cot(A + B) = \cot(\pi - C)$

    $\frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C$

    $\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

    $\Rightarrow \cot^2A + \cot^2B + \cot^2C + 2\cot A\cot B + 2\cot B\cot C + 2\cot C\cot A = 3(\cot A\cot B + \cot B\cot C + \cot C\cot A)$

    $\frac{1}{2}[(\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2] = 0$

    $\Rightarrow \cot A = \cot B = \cot C$

    $\Rightarrow A = B = C$ i.e. triangle is equilateral.

53. Given, $(a^2 + b^2)\sin(A - B) = (a^2 - b^2)\sin(A + B)$

    $(\sin^2A + \sin^2B)\sin(A - B) = (\sin^2A - \sin^2B)\sin(A + B)$

    $[\because \sin^2A - \sin^2B = \sin(A + B)\sin(A - B)]$

    $\Rightarrow (\sin^2A + \sin^2B)\sin(A - B) = \sin^2(A + B)\sin(A - B)$

    $\Rightarrow \sin(A - B)[\sin^2A + \sin^2B - \sin^2C] = 0$

    Either $\sin(A - B) = 0$ or $\sin^2A + \sin^2B - \sin^2C = 0$

    $A = B$ or $a^2 + b^2 - c^2 = 0$

    Thus, triangle is either isosceles or right angled.

54. R.H.S. $= c(\cos A\cos\theta + \sin A\sin\theta) + a(\cos C\cos\theta - \sin C\sin\theta)$

    $= \cos\theta(c\cos A + a\cos C) + \sin\theta(c\sin A - a\sin C)$

    $=b\cos\theta + \sin\theta\left(c.\frac{a}{2R} - a.\frac{c}{2R}\right)$

    $= b\cos\theta =$ L.H.S.

55. The diagram is given below:

    

    $\frac{b}{\sin(B + \theta)} = \frac{a}{2\sin(A - \theta)}$

    $\frac{c}{\sin[\pi -(B + \theta)]} = \frac{a}{2\sin\theta}$

    $\Rightarrow b\sin(A - \theta) = c\sin\theta$

    $\Rightarrow \sin B(\sin A\cos\theta - \cos A\sin\theta) = \sin C\sin\theta = (\sin A\cos B + \sin B \cos A)\sin\theta$

    $\Rightarrow \cot\theta = 2\cot A + \cot B$

56. The diagram is given below:

    

    $\Delta ABC = \Delta ABD + \Delta ACD$

    $\frac{1}{2}bc\sin A = \frac{1}{2}AD.c\sin\frac{A}{2} + \frac{1}{2}AD.b\sin\frac{A}{2}$

    $\Rightarrow AD = \frac{2bc}{b + c}\cos\frac{A}{2}$

57. The diagram is given below:

    

    $\angle ACB = \pi - (\alpha + \beta + \gamma)$

    $\sin ACB = \sin(\alpha + \beta + \gamma)$

    In $\triangle ABC$

    $\frac{AB}{\sin ACB} = \frac{AC}{\sin\gamma}$

    $AC = \frac{a\sin\gamma}{\sin(\alpha + \beta + \gamma)}$

    In $\triangle ACD$

    $\frac{CD}{\sin\alpha} = \frac{AC}{\sin\beta}$

    $CD = \frac{a\sin\alpha\sin\gamma}{\sin\beta\sin(\alpha + \beta + \gamma)}$

58. Given, $2\cos A = \frac{\sin B}{\sin C}$

    $2\cos A\sin C = \sin B = \sin[\pi - (A + C)] = \sin(A + C)$

    $2\cos A\sin C = \sin A\cos C + \cos A\sin C$

    $\cos A\sin C = \sin A\cos C\Rightarrow \tan A = \tan C$

    $\Rightarrow A = C$

    Thus, the triangle is isosceles.

59. Let such angles be $A$ and $B.$ Then,

    $\cos A = \frac{1}{a}$ and $\cos B = \frac{1}{b}$

    $\Rightarrow \sin A\cos A = \sin B\cos B$

    $\sin 2A = \sin 2B$ or $\sin 2A = \sin[\pi - 2B]$

    $A = B$ or $A + B = \frac{\pi}{2} \Rightarrow C = \frac{\pi}{2}$

    Thus, the triangle is either isosceles or right angled.

60. Given $a\tan A + b\tan B = (a + b)\tan \frac{A + B}{2}$

    $\Rightarrow a\left(\frac{\sin A}{\cos A} - \frac{\sin \frac{A + B}{2}}{\cos\frac{A + B}{2}}\right) = b\left(\frac{\sin\frac{A + B}{2}}{\cos \frac{A + B}{2}} - \frac{\sin B}{\cos B}\right)$

    $\Rightarrow a.\frac{\sin\frac{A - B}{2}}{\cos A\cos \frac{A + B}{2}} = b.\frac{\sin\frac{A - B}{2}}{\cos B\cos\frac{A + B}{2}}$

    $\Rightarrow \tan A = \tan B \Rightarrow A = B$

    Thus, the triangle is isosceles.

61. Given, $\frac{\tan A - \tan B}{\tan A + \tan B} = \frac{c - b}{c}$

    $\Rightarrow \frac{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}} = \frac{\sin C - \sin B}{\sin C}$

    $\Rightarrow \frac{\sin A\cos B - \sin B\cos A}{\sin A\cos B + \sin B\cos A} = \frac{\sin (A + B) - \sin B}{\sin (A + B)}$

    $\Rightarrow \frac{\sin A\cos B - \sin B\cos A}{\sin(A + B)} = \frac{\sin(A + B) - \sin B}{\sin(A + B)}$

    $\Rightarrow \sin A\cos B - \sin B\cos A = \sin A\cos B + \sin B\cos A - \sin B$

    $\Rightarrow \sin B = 2\sin B\cos A \Rightarrow \cos A = \frac{1}{2} \Rightarrow A = 60^\circ.$

62. We know that $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

    Given, $c^4 - 2(a^2 + b^2)c^2 + a^4 + a^2b^2 + b^4 = 0$

    $\Rightarrow a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 + 2a^2b^2 = a^2b^2$

    $(a^2 + b^2 -c^2)^2 = a^2b^2 \Rightarrow a^2 + b^2 + c^2 = \pm ab$

    $\Rightarrow \cos C = \pm\frac{1}{2} \Rightarrow A = 60^\circ$ or $120^\circ$

63. Given, $\frac{\cos A + 2\cos C}{\cos A + 2\cos B} = \frac{\sin B}{\sin C}$

    $\Rightarrow \cos A(\sin C - \sin B) = 2\sin B\cos B - 2\sin C\cos C = \sin 2B - \sin 2C$

    $\Rightarrow 2\cos A.\cos\frac{B + C}{2}\sin\frac{C - B}{2} = 2\cos(B + C)\sin(B - C)$

    $\cos A.\cos\frac{B + C}{2}\sin\frac{C - B}{2} = -2\cos A.\sin\frac{B - C}{2}.\cos\frac{B - C}{2}$

    If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles.

    If $A = 90^\circ$ then above is $0 =0$ i.e. triangle is right angled.

64. $\because \tan\frac{A}{2}, \tan\frac{B}{2}, \tan\frac{C}{2}$ are in A.P.

    $\tan\frac{A}{2} - \tan\frac{B}{2} = \tan\frac{B}{2} - \tan \frac{C}{2}$

    $\frac{\sin\left(\frac{A}{2} - \frac{B}{2}\right)}{\cos\frac{A}{2}\cos\frac{B}{2}} = \frac{\sin\left(\frac{B}{2} - \frac{C}{2}\right)}{\cos\frac{B}{2}\cos\frac{C}{2}}$

    $\sin\left(\frac{A}{2} - \frac{B}{2}\right)\cos\frac{C}{2} = \sin\left(\frac{B}{2} - \frac{C}{2}\right)\cos\frac{A}{2}$

    $\sin\left(\frac{A}{2} - \frac{B}{2}\right)\sin\left(\frac{A}{2} + \frac{B}{2}\right) = \sin\left(\frac{B}{2} - \frac{C}{2}\right)\sin\left(\frac{B}{2} + \frac{C}{2}\right)$

    $\Rightarrow \cos B - \cos A = \cos C - \cos B$

    Thus, $\cos A, \cos B, \cos C$ are in A.P.

65. Given, $a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$

    $a.\frac{s(s - c)}{ab} + c.\frac{s(s - a)}{bc} = \frac{3b}{2}$

    $\frac{s}{b}[2s - a - c] = \frac{3b}{2}$

    $2s = 3b \Rightarrow a + c = 2b$

    We have to prove that $\cot\frac{A}{2} + \cot\frac{C}{2} = 2\cot\frac{B}{2}$

    L.H.S. $= \frac{s(s - a)}{\Delta} + \frac{s(s - c)}{\Delta}$

    $= \frac{s}{\Delta}(2s - a - c) = \frac{2s(s - b)}{\Delta} = 2\cot\frac{B}{2} =$ R.H.S.

66. Given, $a^2, b^2, c^2$ are in A.P.

    $\Rightarrow b^2 - a^2 = c^2 - b^2$

    $\Rightarrow \sin^2B - \sin^2A = \sin^2C - \sin^2B$

    $\Rightarrow \sin(B + A)\sin(B - A) = \sin(C + B)\sin(C - B)$

    $\Rightarrow \sin C\sin(B - A) = \sin A\sin(C - B)$

    $\Rightarrow \frac{\sin A\cos B - \cos B\sin A}{\sin A\sin B} = \frac{\sin B\cos C - \sin C\cos B}{\sin B\sin C}$

    $\Rightarrow \cot B - \cot A = \cot C - \cot B$

    $\therefore \cot A, \cot B, \cot C$ are in A.P.

67. Since $A, B, C$ are in A.P. $\Rightarrow 2B = A + C \Rightarrow A + B + C = 3B = 180^\circ \Rightarrow B = 60^\circ$

    Given, $2b^2 = 3c^2$

    $2.\sin^2B = 3.\sin^2C \Rightarrow \sin C = \pm\frac{1}{\sqrt{2}}$

    $\sin C\neq -\frac{1}{\sqrt{2}}$ because $C < 120^\circ$

    $\sin C = \frac{1}{\sqrt{2}} \Rightarrow C = 45^\circ$

    $\Rightarrow A = 75^\circ$

68. Given, $\tan\frac{A}{2}, \tan\frac{B}{2}, \tan\frac{C}{2}$ are in H.P.

    $\Rightarrow \cot \frac{A}{2}, \cot\frac{B}{2}, \cot\frac{C}{2}$ are in A.P.

    $\cot\frac{B}{2} - \cot\frac{A}{2} = \cot\frac{C}{2}- \cot\frac{B}{2}$

    $\frac{s(s - b) - s(s - a)}{\Delta} = \frac{s(s - c) - s(s - b)}{\Delta}$

    $a - b = b - c$

    $a, b, c$ are in A.P.

69. Given, $\frac{\sin A}{\sin C} = \frac{\sin(A - B)}{\sin(B - C)}$

    $\frac{\sin A}{\sin C} = \frac{\sin A\cos B - \sin B \cos A}{\sin B\cos C - \sin C\cos B}$

    $\sin A\sin C\cos C + \sin B\sin C\cos A = 2\sin A\sin C\cos B$

    $\sin B\sin(A + C) = 2\sin A\sin C\cos B$

    $\sin^2B = 2\sin A\sin C\cos B$

    $\cos B = \frac{b^2}{2ac} = \frac{a^2 + c^2 - b^2}{2ac}$

    $\Rightarrow c^2 + a^2 = 2b^2$

    Thus, $a^2, b^2, c^2$ are in A.P.

70. Given, $2\sin B = \sin A + \sin C$

    $4\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A - C}{2} = 2\cos\frac{B}{2}\cos\frac{A - C}{2}$

    $2\sin\frac{B}{2} = 3\cos\frac{A + C}{2} = \cos\frac{A - C}{2}$

    $3\sin\frac{A}{2}\sin\frac{C}{2} = \cos\frac{A}{2}\cos\frac{C}{2}$

    $3\tan\frac{A}{2}\tan\frac{C}{2} = 1$

71. Given, $a^2, b^2, c^2$ are in A.P.

    $b^2 - a^2 = c^2 - b^2$

    $\sin^2B - \sin^2A = \sin^2C - \sin^2B$

    $\sin(A + B)\sin(A - B) = \sin(B + C)\sin(B - C)$

    $\sin C\sin(A - B) = \sin A\sin(B - C)$

    $\cos B -\cot A\sin B = \sin B\cot C - \cos B$

    $2\cos B = \sin B(\cot A + \cot C)$

    $2\cot B = \cot A + \cot C$

    $\therefore \cot A, \cot B, \cot C$ are in A.P.

    $\therefore \tan A,\tan B, \tan C$ are in H.P.

72. We have proven in previous problem that $\therefore \cot A, \cot B, \cot C$ are in A.P.

73. Since $A, B, C$ are in A.P. $\Rightarrow 2B = A + C \Rightarrow A + B + C = 3B = 180^\circ \Rightarrow B = 60^\circ$

    $b:c = \sqrt{3}:\sqrt{2} \Rightarrow \sin C = \frac{\sqrt{3}}{2}.\sqrt{\frac{2}{3}} = \frac{1}{\sqrt{2}}$

    $\Rightarrow C = 45^\circ \Rightarrow A = 75^\circ$

74. Let the sides are $a, b, c$ then $2b = a + c.$ Also, let $a$ to be greatest and $c$ to be smallest side.

    Then, $A = 90 + C$ then $90 + C + B + C = 180 \Rightarrow B = 90 -2C$

    $\frac{a}{\sin(90 + C)} = \frac{b}{\sin(90 - 2C)} = \frac{c}{\sin C} = 2R$

    $4R\cos2C = 2R\cos C + 2R\sin C$

    $2\cos 2C = \cos C + \sin C$

    Squaring, we get

    $4(1 - \sin^22C) = 1 + \sin 2C \Rightarrow \sin2C = \frac{3}{4}$ when $1 + \sin 2C \neq 0$

    When $1 + \sin 2C = 0 \Rightarrow C = \frac{3\pi}{4}$ which is not possible.

    $\because \sin 2C = \frac{3}{4} \Rightarrow \cos 2C = \frac{\sqrt{7}}{4}$

    Now $\sin C$ and $\cos C$ can be found and ratio can be evaluated.

75. $\because a, b, c$ are in A.P. $2b = a + c \Rightarrow a = 2b - c$

    $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{b^2 + c^2 -4b^2 -c^2 +4bc}{2bc}$

    $= \frac{4bc - 3b^2}{2bc} =\frac{4c - 3b}{2c}$

76. The diagram is given below:

    

    Let $AB = 2, AD = 5, BC = 3$ annd $CD=x$

    Since it is cyclic quadrilateral $\angle C = 120^\circ$

    Applying cosine rule in $\Delta ABD,$ we have

    $\cos 60^\circ = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD} \Rightarrow BD^2 = 19$

    Applying cosine rule in $\Delta BCD,$ we have

    $\cos 120^\circ = \frac{BC^2 + CD^2 - BD^2}{2.BC.BD} \Rightarrow x^2 + 3x -10 = 0$

    $x = -5, 2$ but $x$ cannot be -ve. $\therefore x = 2$

77. Given $(a + b + c)(b + c - a) = 3bc$

    $b^2 + c^2 - a^2 + 2bc = 3bc$

    $\frac{b^2 + c^2 - a^2}{2bc} = \frac{1}{2}$

    $\cos A = \cos 60^\circ$

    $A = 60^\circ$

78. Since $AD$ is the median

    $\therefore AB^2 + AC^2 = 2BD^2 + 2AD^2$

    $\Rightarrow b^2 + c^2 = \frac{a^2}{4} + 2AD^2$

    $4AD^2 = b^2 + c^2 + (b^2 + c^2 - a^2)$

    $\cos A = \frac{1}{2} = \frac{b^2 + c^2 - a^2}{2bc}$

    $\Rightarrow 4AD^2 = b^2 + c^2 + bc$

79. The diagram is given below:

    

    Since $AD$ is the median

    $\therefore AB^2 + AC^2 = 2BD^2 + 2AD^2$

    $\Rightarrow AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$

    $AO = \frac{2}{3}AD = \frac{2}{3}.\frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$

    Similalry $BO = \frac{1}{3}\sqrt{2c^2 + 2a^2 - b^2}$

    $\because \angle AOB = 90^\circ$

    $\therefore AO^2 + BO^2 = AB^2$

    $\Rightarrow a^2 + b^2 = 5c^2$

80. Given, $\frac{\tan A}{1} = \frac{\tan B}{2} = \frac{\tan C}{3} = k$

    Since $A, B, C$ are the angles of a triangle

    $\therefore \tan A + \tan B + \tan C = \tan A\tan B\tan C$

    $k + 2k + 3k = k.2k.3k \Rightarrow k = 1$ as if $k = -1$ sum of angles will be greater than $180^\circ.$

    $\tan A = 1 \Rightarrow \sin A = \frac{1}{\sqrt{2}}$

    $\tan A = 2 \Rightarrow \sin A = \frac{2}{\sqrt{5}}$

    $\tan A = 3 \Rightarrow \sin A = \frac{3}{\sqrt{10}}$

    $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

    $\sqrt{2}a = \frac{\sqrt{5}b}{2} = \frac{\sqrt{10}c}{3}$

    $6\sqrt{2}a = 3\sqrt{5}b = 2\sqrt{10}c$

81. For a triangle sides are positive i.e. $a > 0, b > 0, c >0$ where $a,b,c$ are the sides.

    $2x + 1 > 0 \Rightarrow x > -\frac{1}{2}$

    $x^2 - 1>0 \Rightarrow x > 1$ because side cannot be negative.

    $x^2 + x + 1 > 0~\forall x>1$

    $a - b = x(x - 1) > 0 \Rightarrow a > b$

    $a - c = x + 2 > 0 \Rightarrow a > c$

    Hence $a$ is the greatest side.

    $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(2x + 1)^2 + (x^2 - 1)^2 - (x^2 + x + 1)}{2(2x + 1)(x^2 - 1)}$

    $= -\frac{1}{2}$

    $\Rightarrow A = 120^\circ$

82. Let the sides be $x, x+1, x+2$ where $x > 0$ and is a natural number. Let the smallest angle be $\theta$

    $\angle C = \theta \therefore \angle A = 2\theta$

    Applying sine law

    $\frac{x}{\sin\theta} = \frac{x + 1}{\sin(\pi - 3\theta)} = \frac{x + 2}{\sin2\theta}$

    $\Rightarrow \frac{x}{\sin\theta} = \frac{x + 2}{\sin2\theta} \Rightarrow 2\cos\theta = \frac{x + 2}{x}$

    $\Rightarrow \frac{x}{\sin\theta} = \frac{x + 1}{\sin3\theta} = \frac{x + 1}{3\sin\theta - 4\sin^2\theta}$

    $\Rightarrow 3 - 4\sin^2\theta = \frac{x + 1}{x}$

    $\Rightarrow 4\cos^2\theta = \frac{2x + 1}{x} = \frac{(x + 2)^2}{x^2}$

    $\Rightarrow x^2 - 3x - 4 = 0$

    $x = 4, -1$ but $-1$ is not a natural number so $x = 4.$ Hence sides are $4,5,6.$

83. Given, $a = 6$ cm, $\Delta = 12$ sq. cm. and $B - C = 60^\circ$

    $\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}a.k\sin B\sin C$

    $= \frac{1}{2}.a.\frac{a}{\sin A}\sin B\sin C$

    $\Delta = \frac{1}{2}a^2\sin B\sin C = \frac{18\sin B\sin C}{\sin A}$

    $\Rightarrow \frac{2}{3} = \frac{2\sin B\sin C}{2\sin A} = \frac{\cos(B - C) - \cos(B + C)}{2\sin A}$

    $\Rightarrow \frac{2}{3} = \frac{\cos60^\circ - \cos(\pi - A)}{2\sin A}$

    $\Rightarrow 8\sin A - 6\cos A = 3$

84. Given, $\cos\theta = \frac{a}{b + c} \Rightarrow 1 + \cos\theta = \frac{a + b + c}{b + c}$

    $\Rightarrow 2\cos^2\frac{\theta}{2} = \frac{a + b + c}{b + c}$

    $\sec^2\frac{\theta}{2} = \frac{2(b + c)}{a + b + c}$

    $1 + \tan^2\frac{\theta}{2} = \frac{2(b + c)}{a + b + c}$

    Similarly, $1 + \tan^2\frac{\phi}{2} = \frac{2(c + a)}{a + b + c}$

    and $1 + \tan^2\frac{\psi}{2} = \frac{2(a + b)}{a + b + c}$

    Adding, we get $3 + \tan^2\frac{\theta}{2} + \tan^2\frac{\phi}{2} + \tan^2\frac{\psi}{2} = \frac{4(a + b + c)}{a + b + c}$

    $\Rightarrow \tan^2\frac{\theta}{2} + \tan^2\frac{\phi}{2} + \tan^2\frac{\psi}{2} = 1$

85. Since $C$ is the angle of a triangle, $\sin C\leq 1$

    $\therefore \cos A\cos B + \sin A\sin B\geq \cos A\cos B + \sin A\sin B\sin C$

    $\Rightarrow \cos(A - B)\geq 1$

    But $cos(A - B)$ cannot be greater than $1. \therefore \cos(A - B) = 1\Rightarrow A = B$

    Now, $\cos A\cos B + \sin A\sin B\sin C = 1$

    $\Rightarrow \cos^2A + \sin^2A\sin C = 1$

    $\Rightarrow \sin C= 1 \Rightarrow C=90^\circ\Rightarrow A = B = 45^\circ$

    $\Rightarrow a:b:c = \sin A:\sin B: \sin C = 1:1:\sqrt{2}$

86. From the question, $\sin A\sin B\sin C = p$ and $\cos A\cos B\cos C = q$

    $\therefore \tan A\tan B\tan C = \frac{p}{q}$

    Since we are dealing with a triangle $\therefore \tan A + \tan B + \tan C = \tan A\tan B\tan C$

    $\Rightarrow \tan A + \tan B + \tan C = \frac{p}{1}$

    Now, $\tan A\tan B + \tan B\tan C + \tan C\tan A = \frac{\sin A\sin B\cos C + \sin B\sin C\cos A + \sin A\sin C\cos B}{\cos A\cos B\cos C}$

    [ We know that in a triangle $2\sin A\sin B\cos C = \sin^2A + \sin^2B - \sin^2C$ ]

    $\Rightarrow \frac{1}{2q}[(\sin^2A + \sin^2B - \sin^2C) + (\sin^2B + \sin^2C - \sin^2A) + (\sin^2C + \sin^2A - \sin^2B)]$

    $= \frac{1}{2q}[\sin^2 + \sin^2B + \sin^2C]$

    $= \frac{1}{2q}\left[\frac{(1 - \cos2A) + (1 - \cos2B) + (1 - \cos2C)}{2}\right]$

    $= \frac{1}{4q}[4 + 4\cos A\cos B\cos C] = \frac{1 + q}{q}$

    Thus, we see that $\tan A, \tan B, \tan C$ are roots of the given equation.

87. Given, $\sin^3\theta = \sin(A - \theta)\sin(B - \theta)\sin(C - \theta)$

    $4\sin^3\theta = 2\sin(A - \theta)[2\sin(B - \theta)\sin(C - \theta)]$

    $= 2\sin(A - \theta)[\cos(B - C) - \cos(B + C - 2\theta)]$

    $= 2\sin(A - \theta)\cos(B - C) -2\sin(A - \theta)\cos(B + C - 2\theta)$

    $= \sin(A + B - C - \theta) + \sin(A + C - \theta - B) - \sin(A + B + C - 3\theta) + \sin(\pi - 2B - \theta)$

    $\sin 3\theta + 4\sin^3\theta = \sin(2A + \theta) + \sin(2B + \theta) + \sin(2C + \theta)$

    $3\sin\theta = (\sin2A + \sin2B + \sin2C)\cos\theta + (\cos2A + \cos2B + \cos2C)\sin\theta$

    [ $\because \sin2A + \sin2B + \sin 2C = 4\sin A\sin B\sin C$ when $A + B + C = \pi$ ]

    $(1 - \cos 2A) + (1 - \cos 2B) + (1 - \cos 2C)\sin\theta = 4\sin A\sin B\sin C.\cos\theta$

    $2[(\sin^2A + \sin^2B + \sin^2C)\sin\theta] = 4\sin A\sin B\sin C\cos\theta$

    $2\sin\theta[(\sin^2A + \sin^2B - \sin^2C) + (\sin^2B + \sin^2C - \sin^2A) + (\sin^2C + \sin^2A - \sin^2B)] = 4\sin A\sin B\sin C\cos\theta$

    [ $\because \sin^2 + \sin^2B - \sin^2C = 2\sin A\sin B\cos C$ ]

    $\Rightarrow \cot\theta = \cot A + \cot B + \cot C$

88. From question $\frac{b}{c} = r \therefore b = cr$

    Let $AD\perp BC$ and let $AD = h$

    We have to prove that $h\leq \frac{ar}{1 - r^2}$

    $\Delta ABC = \frac{1}{2}c.cr.\sin A = \frac{1}{2}ah$

    $h = \frac{c^2r\sin A}{a}$

    $\cos A = \frac{c^2 + c^2r^2 - a^2}{2.c.cr}$

    $c^2 = \frac{a^2}{1 + r^2 - 2r\cos A}$

    $\Rightarrow h = \frac{a^2r\sin C}{a(1 + r^2 - 2r\cos a)} = \frac{ar\sin A}{1 + r^2 - 2r\cos A}$

    $= \frac{ar.\frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}}}{1 + r^2 - 2r\frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}}$

    $\Rightarrow (1 + r^2)\tan^2\frac{A}{2} -\frac{2ar}{h}\tan\frac{A}{2} + (1 - r)^2 = 0$

    This is a quadratic equation in $\tan\frac{A}{2}$ and since it will be read $D \geq 0$

    $\Rightarrow \frac{4a^2r^2}{h^2} - 4(1 + r)^2(1 - r)^2\geq 0$

    $h \leq \frac{ar}{1 - r^2}$

89. Given $b.c = k^2,$ now

    $\cos A = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow 2k^2\cos A = b^2 + \frac{k^4}{b^2} - a^2$

    $b^4 - (a^2 + 2k^2\cos A)b^2 + k^4 = 0$ which is a quadratic equation in $b^2.$

    The triangle will not exists if discriminant is less than zero for above equation because then $b$ will become a complex number.

    $\Rightarrow (a^2 + 2k^2\cos A)^2 - 4k^4 < 0$

    $\Rightarrow [a^2 + 2k^2(1 + \cos A)][a^2 - 2k^2(1 - \cos A)] < 0$

    $\Rightarrow \left(a^2 + 4k^2\cos^2\frac{A}{2}\right)\left(a^2 - 4k^2\sin^2\frac{A}{3}\right) < 0$

    $\Rightarrow a^2 - 4k^2\sin^2\frac{A}{2} , 0 \left[\because a^2 + 2k^2\cos^2\frac{A}{2} > 0\right]$

    $\Rightarrow \left(a + 2k\sin\frac{A}{2}\right)\left(a - 2k\sin\frac{A}{2}\right) < 0$

    $\Rightarrow a - 2k\sin\frac{A}{2} < 0 \left[\because] a + 2k\sin\frac{A}{2} > 0\right]$

    $\Rightarrow a < 2k\sin\frac{A}{2}$

90. The diagram is given below:

    

    The diagram is a top view. Let $O$ be the top point and $O'$ the center of ring which is $12$ cm below $O$ in the diagram(not shown).

    In triangle $OO'A, AO' = 5$ cm, $OO' = 12$ cm

    $AO = \sqrt{12^2 + 5^2} = 13$ cm

    Now sides of a regular hexagon are equal to the circumscribing circle. $AB= 5$ cm.

    $\cos AOB = \frac{13^2 + 13^2 - 5^2}{2.13.13} = \frac{313}{338}$

91. Given, $2b = 3a$ and $\tan^2\frac{A}{2}= \frac{3}{5}$

    $\cos A = \frac{1}{\sqrt{1 + \tan^2\frac{A}{2}}} = \frac{1}{\sqrt{1 + \frac{3}{5}}} = \sqrt{\frac{5}{8}}$

    $\cos A = \sqrt{\frac{5}{8}} = \frac{b^2 + c^2 - a^2}{2bc}$

    $= \frac{b^2 + c^2 - \frac{4b^2}{9}}{2bc}$

    $\Rightarrow \frac{\sqrt{8}(5b^2 + 9c^2)}{9} = 2\sqrt{5}bc$

    $\Rightarrow 9\sqrt{8}c^2 - 18\sqrt{5}bc + 5\sqrt{8}b^2 = 0$

    $\Rightarrow c = \frac{8\sqrt{5}}{6\sqrt{8}}, \frac{4\sqrt{5}}{6\sqrt{8}}$

    Thus, one value is double of the other.

92. Let the angles are $k, 2k,7k$ degrees. Then $k + 2k + 7k = 180^\circ \Rightarrow k = 18^\circ$

    So greatest angle is $126^\circ$ and smallest is $18^\circ.$

    Ratio of greatest to least side is given by $\sin126^\circ;;\sin18^\circ$

    $= \cos 36^\circ:\sin18^\circ = \sqrt{5} + 1:\sqrt{5} - 1$

93. Let $AF = f, BG = g, CH = h$

    Area of $\triangle ABC =$ Area of $\triangle ABF$ + Area of $\triangle ACF$

    $\frac{1}{2}bc\sin A = \frac{1}{2}.2\sin\frac{A}{2}\cos\frac{A}{2} = \frac{1}{2cf}\sin\frac{A}{2} + \frac{1}{2}bf\sin\frac{A}{2}$

    $\Rightarrow 2bc\cos\frac{A}{2} = (b + c)f$

    $\frac{1}{f}\cos\frac{A}{2} = \frac{1}{2}\left(\frac{1}{b} + \frac{1}{c}\right)$

    Similarly, $\frac{1}{g}\cos\frac{B}{2} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{c}\right)$

    and, $\frac{1}{h}\cos\frac{C}{2} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right)$

    Adding these three we obtain desired result.

94. Since $BD = DE = EC$ each will be equal to $\frac{5}{3}.$ Clearly the triangle is right angled because $3^2 + 5^2 = 5^2$

    $\cos C = \frac{4}{5}$

    In $\triangle ACE, \cos C = \frac{CE^2 + 4^2 - AE^2}{2.CE.4} = \frac{\frac{25}{9} + 16 - AE^2}{2.\frac{5}{3}.4}$

    $\Rightarrow \frac{169 - 9AE^2}{9}.\frac{3}{40} = \frac{4}{5}$

    $\Rightarrow AE^2 = \frac{73}{9}$

    $\cos\theta = \frac{AE^2 + AC^2 - CE^2}{2.AE.AC} = \frac{8}{\sqrt{73}}$

    $\tan\theta = \sqrt{sec^2\theta - 1} = \sqrt{\frac{73}{64} - 1} = \frac{3}{8}$

95. Let $O$ be the centroid i.e. point of intersection of medians.

    From geometry, we know that area of $\triangle ABC = 3\times$ area of $\triangle AOC$

    We also know that centroid divides median in the ratio of $2:1$ i.e. $AO = \frac{10}{3}$

    Applying sine rule in $\triangle AOC$

    $\frac{OC}{\sin\frac{\pi}{8}} = \frac{AO}{\sin\frac{\pi}{4}}$

    $OC = \frac{10}{3}\frac{\sin\frac{\pi}{8}}{\sin\frac{\pi}{4}}$

    Area of $\triangle AOC = \frac{1}{2}AO.OC\sin AOC$

    $= \frac{1}{2}\frac{10}{3}.\frac{10}{3}.\frac{\sin\frac{\pi}{8}}{\sin\frac{\pi}{4}}\sin\left(\frac{\pi}{2} + \frac{\pi}{8}\right)$

    $= \frac{25}{9}$

    $\therefore \Delta ABC = \frac{75}{9}.$

96. Let sides are $a, b, c$ then $a = 7 = \sqrt{49}$ cm, $b = 4\sqrt{3} = \sqrt{48}$ cm and $c = \sqrt{13}$ cm.

    Clearly, $c$ is smallest and thus $C$ will be smallest.

    $\cos C = \frac{48 + 49 - 13}{2.7.4\sqrt{3}} = \frac{3}{2}$

    $\Rightarrow C = 30^\circ$

97. Let the triangle be $ABC$ having right angle at $C.$ Let $D$ be the mid-point of $AC.$

    Given that triangle is isoceles so $AC = BC$ i.e. $DC = \frac{1}{2}AC = \frac{1}{2}BC$

    Also, $\angle CAB = \angle BDA = 45^\circ$

    Let $\angle DBC = \theta$ and $\angle DBA = \phi$

    $\tan\phi = \frac{DC}{BC}= \frac{1}{2}$

    $\tan\phi = tan(45^\circ - \theta) = \frac{1 - \tan\theta}{1 + \tan\theta}$

    $\Rightarrow \tan\phi = \frac{1}{3}$

    $\therefore \cot\theta = 2, \cot\phi = 3$

98. From the given ratios we have,

    $\frac{a + b}{(1 + m^2)(1 + n^2)}= \frac{a - b}{(1 - m^2)(1 - n^2)} = \frac{c}{(1 - m^2)(1 + n^2)}$

    $\Rightarrow \frac{a + b }{c} = \frac{1 + m^2}{1 - m^2}, \frac{a - b}{c} = \frac{1 - n^2}{1 + n^2}$

    $\Rightarrow \frac{\cos\frac{A - B}{2}}{\cos\frac{A + B}{2}} = \frac{1 + m^2}{1 - m^2}, \frac{\sin\frac{A - B}{2}}{\sin\frac{A + B}{2}} = \frac{1 - n^2}{1 + n^2}$

    By componendo and dvidendo, we have

    $\tan\frac{A}{2}\tan\frac{B}{2} = m^2, \cot\frac{A}{2}\tan\frac{B}{2} = n^2$

    $\Rightarrow \tan^2\frac{A}{2} = \frac{m^2}{n^2}, \tan^2\frac{B}{2} = m^2n^2$

    $\Rightarrow A = 2\tan^{-1}\frac{m}{n}, B = 2\tan^{-1}mn$

    $\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}bc\frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}}$

    $= \frac{mnbc}{m^2 + n^2}$

99. Since $a,b,c$ are roots of the equation $x^3 - px^2 + qx - r = 0$ therefore we have

    $a + b + c = p = 2s$ where $s$ is perimeter.

    $ab + bc + ca = q$ and $abc = r$

    $\Delta^2 = s(s- a)(s - b)(s - c)$

    $= \frac{p}{2}\left(\frac{p}{2} -a\right)\left(\frac{p}{2} - b\right)\left(\frac{p}{2} - c\right)$

    Substituting the values of we obtain the desired result.

100. Let the third side be $a$ cm. Applying cosine rule,

     $6 = a^2 + 4^2 - 2.a.4\cos30^\circ$

     $a^2 - 4\sqrt{3}a + 10 = 0$

     $a = \frac{4\sqrt{3}\pm\sqrt{48 - 40}}{2} = 2\sqrt{3} \pm \sqrt{2}$

     Both roots are positive, so two such triangles are possible.